Question

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of \(^{63}_{ 29}Cu\) atoms (of mass 62.92960 u).

Answer 1

Mass of a copper coin, \(m’ = 3 g\)
Atomic mass of \(^{63}_{ 29}Cu\) atom, m = 62.92960 u
The total number of \(^{63}_{ 29}Cu\) atoms in the coin
N = \(\frac{NA \times  m'}{Mass Number}\)
Where, 
NA = Avogadro’s number = 6.023 × 10²³ atoms /g 
Mass number = 63 g
\(N =\frac{6.023 \times  10^{23} \times  3 }{63} = 2.868 \times  10^{22} atoms\)
\(^{63}_{ 29}Cu\) nucleus has 29 protons and (63 − 29) 34 neutrons
Mass defect of this nucleus, ∆m' = 29 × m_H + 34 × m_n − m 
Where, 
Mass of a proton, m_H = 1.007825 u 
Mass of a neutron, m_n = 1.008665 u 
∆m' = 29 × 1.007825 + 34 × 1.008665 − 62.9296 
= 0.591935 u 
Mass defect of all the atoms present in the coin, 
∆m = 0.591935 × 2.868 × 10²² = 1.69766958 × 10²² u 
But 1 u = 931.5 \(\frac{MeV}{c^2} \)
∆m = 1.69766958 × 10²² × 931.5 \(\frac{MeV}{c^2} \)
Hence, the binding energy of the nuclei of the coin is given as: 
\(E_b= ∆mc^2 \)
= 1.69766958 × 10²² × 931.5 \((\frac{MeV}{c^2})c^2\)
= 1.581 × 10²⁵ MeV 
But 1 MeV = 1.6 × 10⁻¹³ J 
E_b = 1.581 × 10²⁵ × 1.6 × 10⁻¹³ = 2.5296 × 10¹² J 
This much energy is required to separate all the neutrons and protons from the given coin.